5.2.1
Notations
We use the following notations throughout this chapter. Calligraphic letters are used to represent sets and their corresponding capital letters are used to represent the cardinalities
of these set. Let N be a set. Then, N denotes the cardinality of N, i.e., |N |=N. Further,
P(N) represents the power set ofN. The set denoted by M × N represents the Cartesian product of the two sets Mand N.
5.2.2
System Model
We consider a network model in which M geographically close devices are interested in receiving N packets. Let M = {U1, ..., UM} denote the set of devices wherein each device
Ui ∈ M initially holds a subset of packets from N = {P1, ..., PN}. Such prior received
packets model the side information of devices or received packets from the base station in previous broadcast sessions [9, 25]. Moreover, we consider order constrained applications for which packets can only be delivered and used at the application layer sequentially in order. In other words, packet Pj is delivered to the application layer of device Ui only if packets
P1, · · · , Pj are decoded. Out-of-order decoded packets are stored in the devices’ buffers,
but cannot be delivered to the application layer.
We consider that all devices are equipped with two wireless interfaces. In other words, devices are able to use long-range cellular and short-range local area channels concurrently for transmission and reception. We herein assume that all devices are in the transmission range of each other forming a fully connected cooperative network. Hence, to avoid interference, a single device is allowed to transmit in the D2D communications spectrum in each time slot. Therefore, in addition to an XORed packet transmission from the base station, one of theM devices simultaneously transmits an XORed packet using a different part of the spectrum. Such protocol may result in receiving either a single or two packets by each device at a time slot t.
Each transmitted packet from device Ui to device Uk is subject to an independent
Bernoulli erasure with probability ǫi,k. For notational simplicity, we denote the base station
by U0 with ǫ0,k representing its channel erasure probability to device Uk. The channels are
independent, but not necessarily identical in terms of erasure probability and their statistics are assumed to be known to the base station.
Similar to Chapters 3 and 4, in this chapter, we consider a centralized decision making system to solve the in-order packet delivery problem in a heterogeneous network for the following advantages. First, a centralized system satisfies and executes reliably the service
providers policy of quick in-order packet delivery to all devices. Second, it can optimally solve the quick in-order packet delivery problem using the high computational capabilities of the base station. Third, the centralized implementation has low processing requirements at the devices. Finally, it is adaptive to the devices’ mobility as the base station can deploy additional resources to be adaptive to the variations in the network [31].
At each time slott, the base station selects an XOR packet combination to be transmitted by itself, a transmitting device and an XOR packet combination for the transmitting device. The decision is made using the information about the diversity of lost and received packets at the devices as well as the channels’ erasure probabilities. Such information assemblage is accomplished by the collection of two feedback bits per device (i.e., one bit for each interface) after each time slot. After the reception of the feedback bits, the base station updates the information about the packet reception status of all devices and stores it in an M × N
feedback status matrix (FSM) F= [fk,l], ∀(Uk, Pl)∈ M × N, such that:
fk,l =
0 if packet Pl is decoded by device Uk,
1 if packet Pl is missing at device Uk.
(5.1)
Example 11. An example of FSM for a network composed of M = 2 devices and N = 4
packets is given as follows:
F= 1 0 1 0 0 0 1 1 . (5.2)
Based on the reception status of the packets at a given time slot, the following three sets are attributed to each device Uk:
• The Has set Hk is defined as the set of packets successfully decoded by device Uk. In
Example 11, the Has set of device U1 is H1 ={P2, P4}.
• The Wants set Wk =N \ Hk is defined as the set of missing packets at device Uk. In
Example 11, the Wants set of device U1 is W1 ={P1, P3}.
• The Delivered set Lk ⊆ Hk is defined as the set of packets delivered to the application
packet. In Example 11, the Delivered set of devices U1 and U2 are L1 = ∅ and
L2 = {P1, P2}, respectively. Therefore, the undelivered set of devices U1 and U2 are
N \ L1 ={P1, P2, P3, P4} and N \ L2 ={P3, P4}, respectively.
Throughout this chapter, the notation We
k ∈ Wk denotes the index of the e-th missing
packet of device Uk. In Example 11,W21 = 3 refers to the first missing packet of device U2,
i.e., packet P3, andW22 = 4 refers to the second missing packet of device U2, i.e., packet P4.
5.2.3
Delay Metrics
In this subsection, we introduce the relevant definitions used throughout this chapter. We first define instantly decodable transmissions as follows:
Definition 19 (Instantly Decodable Transmission). A transmitted packet combination is instantly decodable for device Uk if it contains exactly one packet from its Wants set Wk [9, 52, 63, 93].
We now introduce different delay metrics of IDNC-enabled networks. The completion time is widely regarded as an appropriate metric to quantify the throughput of IDNC-enabled systems. Note that the completion time metric is inversely proportional to throughput and can be converted into throughput with the usage of bandwidth and packet size. The completion time metric, denoted by τ, is defined as follows:
Definition 20 (Completion Time). The completion time τ is defined as the number of time slots required until all M devices recover all N packets [93].
Example 12. Consider the FSM F in Example 11 and assume erasure-free transmissions. Let S be the schedule of transmissions in which:
1. The base station and device U2 send packetP3 and packet P1, respectively, in the first
time slot.
2. The base station or device U1 broadcasts packet P4 in the second time slot.
The evolution of the FSM after each time slot is given by: 1 0 1 0 0 0 1 1 ⇒ t=1 0 0 0 0 0 0 0 1 ⇒ t=2 0 0 0 0 0 0 0 0 . (5.3)
The schedule S requires two time slots to complete the reception of all packets by all devices. Therefore, the completion time is τ = 2.
Despite its direct impact on the throughput, the completion time metric is not suitable for order-constrained applications since it only considers the minimum number of time slots regardless of the order of transmitted packets. For example, the reversed transmissions in Example 12 yields the same completion time but different patterns of in-order delivered packets. A more appropriate metric for order-constrained applications is the delivery time, which can be defined as follows:
Definition 21 (Individual Delivery Time). The individual delivery time Tk of device Uk increases by one unit for each undelivered packet in each transmission. In other words, the individual delivery time of device Uk increases by |N \ Lk|=N −Wk1+ 1 units for|N \ Lk| undelivered packets in every time slot before recovering all N packets.
Definition 22 (Overall Delivery Time). The overall delivery time T is the summation of the individual delivery times of all devices over all the time slots until the reception of all N
packets by all M devices.
Example 13. To illustrate the delivery time metric, consider the evolution of FSM in (5.3)
using the schedule S described in Example 12. After the first time slot, the only undelivered packet is packet P4 of device U2. Such schedule results in the overall delivery time T = 1.
Let S′ be the schedule in which the order of the transmissions are reversed. The evolution of
the FSM after each time slot is given by: 1 0 1 0 0 0 1 1 ⇒ t=1 1 0 1 0 0 0 1 0 ⇒ t=2 0 0 0 0 0 0 0 0 . (5.4)
After the first time slot, the undelivered packets of device U1 are P1, P2, P3, P4 and the un-
delivered packets of device U2 are P3, P4. Such schedule results in the overall delivery time
T = 4 + 2 = 6. Note that both schedules have the same completion timeτ = 2 but experience very different delivery times.